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#!/usr/local/bin/perl
use strict;
use warnings;
use feature qw(say);
use Test::More;
is( is_palindrome_rev(1221), 1 ); ## Even number of digits
is( is_palindrome_rev(-101), 0 ); ## -ve number
is( is_palindrome_rev(101), 1 ); ## Odd number of digits
is( is_palindrome_rev(90), 0 ); ## Not-palindromic
is( is_palindrome_array(1221), 1 );
is( is_palindrome_array(-101), 0 );
is( is_palindrome_array(101), 1 );
is( is_palindrome_array(90), 0 );
done_testing();
say is_palindrome_array($_,2), is_palindrome_array($_,3), is_palindrome_array($_,4), is_palindrome_array($_,6),' ',$_ foreach 6..1000000;
## I will provide two solutions here...
## (1) The first one just treating the two numbers as strings and
## using reverse
sub is_palindrome_rev {
return ( $_[0] eq reverse $_[0]) ? 1 : 0;
}
## (2) The second one just treating the two numbers as numbers and
## using the % operator to split them into digits.
## Notes:
## * Push is more optimal than unshift - so the digit list I get
## is in reverse order but that doesn't matter in this case
## * Obviously -ve numbers will fail so deal with this first
## * When the digits array has one element then we know the
## remaining digits form a palindrome {we can't use the
## shift/pop trick as it will compare digit with undef;
sub is_palindrome_array {
my($n,$radix) = @_;
$radix||=10;
return 0 if $n < 0;
my @digits = $n%$radix;
push @digits, $n%$radix while $n = int ($n/$radix);
while( @digits>1 ) {
return 0 if shift @digits != pop @digits;
}
return 1;
}
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